Modern Engineering Thermodynamics

9.9 Unsteady State Processes in Open Systems 307
Assuming the flow to be incompressible and adiabatic, the entropy production rate is given by Eq. (9.53) as
_S P = ð500: lbm/sÞ½1:00 Btu/ ðlbm.RÞŠ
2
3
6
ð32:174 ft/s 2 Þð0:0711 ftÞ
7
× ln 41 +
1:00 Btu
32:174 lbm
5
.ft
lbm.R lbf .s 2 ð50:0 + 459:67 RÞ
= 0:0697 Btu/ ðs.RÞ
Exercises
26. Determine the entropy production rate in the hydraulic jump discussed in Example 9.10 if the mass flow rate of the
water is reduced from 500. lbm/s to 50.0 lbm/s. Keep the values of all the other variables the same as they are in
Example 9.10. Answer: _S P = 0:00697 Btu/(s·R).
27. If the exit water height (y 2 ) in Example 9.10 is 1.50 ft rather than 1.80 ft, what is the entropy production rate of the
hydraulic jump. Keep the values of all the other variables the same as they are in Example 9.10. Answer: _S P = 0:0204
Btu/(s·R).
28. If the inlet water temperature in Example 9.10 is increased from 50.0°F to 80.0°F, what is the new entropy production
rate of the hydraulic jump? Keep the values of all the other variables the same as they are in Example 9.10. Answer:
_S P = 0:0659 Btu/(s·R).
The next example illustrates the use of the more complex direct method of determining the entropy production
rate in a simple laminar hydrodynamic flow situation. In the design of hydraulic systems, the viscous losses in
the flow stream oil can be very large. For this reason, it is usually desirable to keep the flow of hydraulic oil
laminar rather than turbulent.
EXAMPLE 9.11
The velocity profile in the steady isothermal laminar flow of an incompressible Newtonian fluid in a horizontal circular tube
of radius R is given by
V = V m 1 − ðx/RÞ 2
where V m is the maximum (i.e., centerline) velocity of the fluid, and x is the radial coordinate measured from the centerline
of the tube.
Determine the entropy production rate due to laminar viscous losses in water at 20.0°C flowing in a 2.50 cm diameter pipe
with a centerline velocity of 0.500 m/s. The viscosity of the water is 10.1 × 10 –3 kg/(m· s), and the length of the pipe is
10.0 m.
Solution
First, draw a sketch of the system (Figure 9.22).
The unknown is the entropy production rate due to laminar viscous losses. The material is liquid water.
Here we use Eq. (7.72), which is the direct method, to determine the
entropy production rate due to fluid viscosity. By differentiating
the velocity formula given in the problem statement, we get
then
dV
dx = −2V m x/R 2
x
Annular
differential
volume
dx
and Eq. (7.72) gives
ðσ W
Þ vis
= μ
T
Z
_S P
W =
vis
dV
dx
2 4μV 2
=
m x2
R 4 T
Vðσ W Þ vis
dV
FIGURE 9.22
Example 9.11.
L
(S P ) W-vis = ?
(Continued )